# Question 90c2e

Sep 29, 2016

$3$

#### Explanation:

Given

${L}_{1} \to p = {p}_{1} + {\lambda}_{1} {\vec{v}}_{1}$ and
${L}_{2} \to p = {p}_{2} + {\lambda}_{2} {\vec{v}}_{2}$

with

${p}_{1} = \left(1 , - 3 , 0\right)$
${\vec{v}}_{1} = \left(4 , 1 , - 1\right)$
${p}_{2} = \left(1 , 1 , 1\right)$
${\vec{v}}_{2} = \left(2 , 0 , - 1\right)$

The sought plane $\Pi$ contains ${L}_{1}$ and does not contain ${L}_{2}$, being parallel to it, so their distance is constant.

Then

$\Pi \to p = {p}_{1} + {\lambda}_{1} {\vec{v}}_{1} + {\lambda}_{2} {\vec{v}}_{2}$

And ${p}_{0}$ the ${p}_{2}$ projection onto $\Pi$ is the solution for

{( p_0 = p_1 + lambda_1 vec v_1 + lambda_2 vec v_2),( << p_2-p_0, vec v_1 >> = 0),( << p_2-p_0, vec v_2 >> = 0):}#

Giving ${\lambda}_{1} = \frac{8}{3} , {\lambda}_{2} = - 5$

so ${p}_{0} = \left(\frac{5}{3} , - \frac{1}{3} , \frac{7}{3}\right)$ and the distance $d = \left\lVert {p}_{2} - {p}_{0} \right\rVert = 2$